LW5 Lecture Notes

MAT A31 2014 Lecture Week 5 Textbook: Salas, Hille, Etgen: CALCULUS, One Variable, 10e 5.1 Implicit Differentiation Implicit Differentiation A function is defined explicitly if it can be described by expressing one variable in term of another. For example: - A fimction y is defined implicitly if it is described by a relation between z and y.. Fé_r example: b ] 2 ; 'y ='x:"' —~4x x84 8 = dx2y? —3y2xt — 3x2y4 sin{(m(x 4+ ) =0 an elliptic curve a four-leaved rose If the function y is given implicitly, to find the derivative of y we use implicit differentiation. This consists of differentiating both sides of the equation with respect to z and then solving the resulting equation for y/. — / . ) Example: If 2? + 2y —y® =4 find % l\(?(y()());,:gfl/ y;' =£J/' 3//3% [x"#yy—yjj':(/f)' vl C2y?) Yo -y Ar+(y7 3% )3y =0 (x-34) 3= | %i . Y+ Y L dy -(Ax1%) | £XT Ly s J-sff fo el I Homework: If cos(z +y) =y? find %. . bateg) ()" el -sinony "&l"'('"w'('(*'w': Y dy-. - sin(x+¥) ~gin (x+ Wflf%;xfi&%{% i gan(iiy)t Y _ Jq'n(Y-t y) = #n (x+y) %*%% 1

Derivatives of Inverse Ifunctions Derivatives of Inverse Trigonometric Functions D 1 arcsin' z = i for z € (—1,1). » Recall: B " y =arcsinz <= siny=z and -3 <y < ) and ~1<z<1 We use implicit differentiation on siny =z . . ' , (fi'ng):(x/ (! ! F/:] wdyay =./ =7 g'my |}/ 47"% |)7 XL'/i Jloz Using the same technique we can get the following: ) 1 S —1 arcsin'' ¢ = Arcesc e = l—gz ay/zr =1 arccos T = : BLCBECT = T . 1 -1 arctan'm = S arccot''s = A 23+ 1 142 Homework: 1)Show: arccos' « = —\/I: for z € (—1,1). Prfncipa[ vokee 9} 203 X i [0 'fjln ferv. Ay A — 1 -1 = e — - dw &ny ey VI-x 2) Find % ify= arccos(q,'?)earcsin:c Qresan X ',1(% [a/rcwd[x)]/ el + arccod(X*) 4 o [a/rc:/rn)() oy -1 orcsin X L arcank f %Y Ax-e 4 arccos(c') L ' ax W {—X*

tog, X x 2 X=gfi'x X=4 W a0 0 Derivatives of the Logarithmic Function. T and in particular, (Inz) = 51;- ,{(77'- x=dd x'=(a 'f"?')@ 1= b4 (49}/')':7({%9/: ! '[i'fi X (log,z) = Xa m 5 . find f'(z). Homework: If f(z) = ~lh ,]l(x) Xt¥ W) X 'L(X'5)'/X+?)'VY'5 g—_—_g—___ VXx-5 (W?L Jx-5" (x+y)2 "(x—5)()rf¥/ 5.2 Logarithmic Differentiation. E%(Mny) 4 (s1 X) Example: Let . N (z—1)°(z — )" fo) = G oy + 5 £ 6 Calculate f'(z). SOLUTION. /fi,, 'fi(x) 5%{(x /) + flu(X- 6) -2t [X"o?.) jfif (Yz-ffl)f'fé) %%{x/) (3t (x~ n)" (m[x 5)) (2 bu(x 2))' @fl((x+1x+ 5)) 72[(5)()) [Xsl xllg X4 y'+,,1x+ 6} 7[;() £(x): { }] Steps in Logarithm Differentiation 1. Take natural logarithms of both sides of the equation ¥ = f(z) and use the Laws of Lotafithms to simplify. 2. Differentiate implicitly with respect to z. 3. Solve the resulting equation for /.